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Sagot :
Let's analyze the growth data for plants in soils A and B to determine which soil tended to have greater growth.
First, we calculate the mean growth for each soil type:
- Mean growth for Soil A: [tex]\(4.17\)[/tex] cm
- Mean growth for Soil B: [tex]\(4.28\)[/tex] cm
Next, we calculate the standard deviation for each soil type, which measures the variation or dispersion of the growth data:
- Standard deviation for Soil A: [tex]\(0.8391\)[/tex] cm
- Standard deviation for Soil B: [tex]\(0.6925\)[/tex] cm
We also calculate the median growth for each soil type, which is the middle value when the data points are ordered:
- Median growth for Soil A: [tex]\(3.9\)[/tex] cm
- Median growth for Soil B: [tex]\(4.25\)[/tex] cm
Finally, we calculate the interquartile range (IQR), which is the difference between the 75th percentile and the 25th percentile, providing a measure of statistical dispersion:
- IQR for Soil A: [tex]\(0.975\)[/tex] cm
- IQR for Soil B: [tex]\(0.825\)[/tex] cm
Now, we assess the given statements:
1. "Soil A had greater growth because its data have a greater standard deviation."
- This statement is not accurate since standard deviation alone does not determine greater growth; it only indicates variability in the data.
2. "Soil A had greater growth because its data have a greater mean."
- This statement is incorrect because the mean growth for Soil A ([tex]\(4.17\)[/tex] cm) is actually less than that for Soil B ([tex]\(4.28\)[/tex] cm).
3. "Soil B had greater growth because its data have a greater interquartile range."
- This statement is incorrect because Soil B has a smaller IQR ([tex]\(0.825\)[/tex] cm) compared to Soil A ([tex]\(0.975\)[/tex] cm).
4. "Soil B had greater growth because its data have a greater median."
- This statement is correct because the median growth for Soil B ([tex]\(4.25\)[/tex] cm) is greater than the median growth for Soil A ([tex]\(3.9\)[/tex] cm).
Therefore, the correct statement is:
Soil B had greater growth because its data have a greater median.
First, we calculate the mean growth for each soil type:
- Mean growth for Soil A: [tex]\(4.17\)[/tex] cm
- Mean growth for Soil B: [tex]\(4.28\)[/tex] cm
Next, we calculate the standard deviation for each soil type, which measures the variation or dispersion of the growth data:
- Standard deviation for Soil A: [tex]\(0.8391\)[/tex] cm
- Standard deviation for Soil B: [tex]\(0.6925\)[/tex] cm
We also calculate the median growth for each soil type, which is the middle value when the data points are ordered:
- Median growth for Soil A: [tex]\(3.9\)[/tex] cm
- Median growth for Soil B: [tex]\(4.25\)[/tex] cm
Finally, we calculate the interquartile range (IQR), which is the difference between the 75th percentile and the 25th percentile, providing a measure of statistical dispersion:
- IQR for Soil A: [tex]\(0.975\)[/tex] cm
- IQR for Soil B: [tex]\(0.825\)[/tex] cm
Now, we assess the given statements:
1. "Soil A had greater growth because its data have a greater standard deviation."
- This statement is not accurate since standard deviation alone does not determine greater growth; it only indicates variability in the data.
2. "Soil A had greater growth because its data have a greater mean."
- This statement is incorrect because the mean growth for Soil A ([tex]\(4.17\)[/tex] cm) is actually less than that for Soil B ([tex]\(4.28\)[/tex] cm).
3. "Soil B had greater growth because its data have a greater interquartile range."
- This statement is incorrect because Soil B has a smaller IQR ([tex]\(0.825\)[/tex] cm) compared to Soil A ([tex]\(0.975\)[/tex] cm).
4. "Soil B had greater growth because its data have a greater median."
- This statement is correct because the median growth for Soil B ([tex]\(4.25\)[/tex] cm) is greater than the median growth for Soil A ([tex]\(3.9\)[/tex] cm).
Therefore, the correct statement is:
Soil B had greater growth because its data have a greater median.
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